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Let $V$ be a finite dimensional vector space and $T \in \mathcal L (V).$ Then for any $\lambda, z \in \mathbb C$ consider the following identity $:$ $$-(T - \lambda I) = (z I - T) - (z - \lambda) I.$$

This shows that $\lambda$ is an eigenvalue of $T$ if and only if $(z - \lambda)$ is an eigenvalue of $(z I - T).$ Raising both sides of the above equation to the $\dim V$ power and taking null spaces of both sides shows that the multiplicity of $\lambda$ as an eigenvalue of $T$ coincides with the multiplicity of $(z - \lambda)$ as an eigenvalue of $(z I - T).$ The bold faced argument is given in Sheldon Axler's book Linear Algebra Done Right which I am unable to follow. Could anyone please provide some insight on it?

Thanks for your time.

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    $\begingroup$ You are basically just translating the spectrum of your matrix when you add a multiple of the identity $\endgroup$
    – Lelouch
    Commented Sep 9, 2022 at 14:45

2 Answers 2

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The desired result follows from the sentence above in boldface and the book's definition of the multiplicity of $\lambda$ as an eigenvalue of $T$ to be dim null $(T - \lambda I)^{\dim V}$.

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  • $\begingroup$ Now I get it. Thanks sir. $\endgroup$ Commented Sep 11, 2022 at 3:09
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Let $n=\dim V.$ The multiplicity of the eigenvalue $\lambda$ of $T$ is by definition equal to $\dim\ker(\lambda I-T)^n.$ We have $$\dim\ker (\lambda I-T)^n=\dim\ker [(z-\lambda)I-(zI-T)]^n$$ Therefore a number $\lambda_0$ is an eigenvalue with multiplicity $k$ of $T$ if and only if $z-\lambda_0$ is an eigenvalue with multiplicity $k$ of $zI-T.$

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  • $\begingroup$ Sheldon Axler proved as a corollary to the argument (I mentioned in bold faced letters) that the characteristic polynomial of $T$ is of the form $\det (\lambda I - T).$ So we can't use this to prove the result what we are supposed to. You didn't use the anywhere the author's line of reasoning. I just want you to clarify only the bold faced argument if you can. $\endgroup$ Commented Sep 9, 2022 at 14:27
  • $\begingroup$ How do you know that the algebraic multiplicity of $\lambda$ equals to $\dim \text {ker} (\lambda I - T)^n\ $? $\endgroup$ Commented Sep 9, 2022 at 17:36
  • $\begingroup$ Why don't you ask @Sheldon Axler the same question as in your second comment ? $\endgroup$ Commented Sep 11, 2022 at 17:29

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